Quantum Key Distribution - cse.iitk.ac.in€¦ · Quantum cryptography makes use of laws of quantum...

Quantum Key Distribution

Abhinandan ShuklaNaman Gupta

November 16, 2017

Abhinandan Shukla Naman Gupta QKD November 16, 2017 1 / 21

Background

Cryptography is the method of performing communication whilekeeping this transferred information safe from any adversary.

Quantum cryptography makes use of laws of quantum mechanics toperform cryptographic tasks.

A lot of cryptographic tasks is based on encrypting and decryptingthe message to be sent using a key, which can be public or private.

Quantum Key Distribution protocols allow us to securely generate asecret key.


Background

Some of the most famous QKD protocols are BB84 and E91.

BB84 protocol uses pulses of polarized light, which are in conjugatebasis, to transfer the key.

The E91 protocol, that makes use of the properties of entanglementto generate a random key and detect if any attempt of eavesdroppinghas been made or not.


Our Project

In our project, we are trying to analyze the quantum key distributionprotocols used in different circumstances which are: absence of classicalcommunication, attack by a postquantum eavesdropper limited only by theimpossibility of superluminal signaling and a introduction of noise inpresence of postquantum eavesdropper.


QKD without classical communication

Generally all the QKD protocols use both classical and quantumchannel of communication to ensure that it has generated and shareda secret key securely.

In this protocol proposed by Xiaoyu Li in 2002, we make use ofGreenburger-Horne-Zeilinger(GHZ) states to perform QKD withoutthe help of classical channel.


Protocol

We have two agents, Alice and Bob, who secretly want to share a key.

Alice creates n GHZ states |∆ > , and sends the third qubit to Bob.

A GHZ state is an entangled state of three qubit system expressed as:

|∆ >=1√2

(|000 > +|111 >)


Protocol

When Bob receives the qubits, he performs σx operation or doesnothing at random on each qubit. When he performs σx , he writesdown 1 while when he does nothing, he writes down 0. Finally he hasa n-bit string k.

σx =

[0 11 0

]If we perform a unitary operation σx on the third qubit of the GHZstate, then we get:

|∆′ >=1√2

(|001 > +|110 >)


Protocol

Now Bob sends the qubits back to Alice.

On receiving the qubits, Alice combines these bits with thecorresponding two bits and then performs CNOT on first and secondqubit of every tripartite system( three qubits system) with first one asthe target bit and second one as the control bit.

If Bob hadn’t applied σx to a particular qubit then performing CNOTwill result in :

1√2

(|000 > +|011 >) = |0 > ⊗ 1√2

(|00 > +|11 >)

While if Bob had applied σx to a particular qubit then performingCNOT will result in :

1√2

(|001 > +|010 >) = |0 > ⊗ 1√2

(|01 > +|10 >)


Protocol

Define

|Φ+ >=1√2

(|00 > +|11 >)

|Φ− >=1√2

(|00 > −|11 >)

|Ψ+ >=1√2

(|01 > +|10 >)

|Ψ− >=1√2

(|01 > −|10 >)

Now Alice does a Bell state measurement on the last two qubits.When the measurement outcome is |Φ+ >, she writes down 0 andwhen it is |Ψ+ >, she writes down 1.

But if the measurement outcome is |Φ− > or |Ψ− >, the protocolfails. So Alice abandons it and creates new n GHZ states and repeatsthe protocol


Protocol

At last Alice also gets a n-bit string k' and she can assure k' = k. Itis the key shared by both the sides.

If Bob receives the qubits again then it means that the protocol hasfailed and he begins to establish a new key. We can define a maximumnumber of failed attempts, say 10, after which they abort the process.

We define time interval, tc , such that if Bob hasnt received the qubitsresent from Alice after tc , he knows that the key has been establishedsuccessfully


Security of the quantum key distribution protcol

The security is ensured by the fundamental principle of quantummechanics pertaining to No-cloning of states, whatever Eve does tointercerpt the key will inevitably make her existence known. We statethe proof of it below.

It is important to note that Eve can intercept and capture the qubitswhen Alice sends them to Bob or vice-versa. But its useless for Eve tocatch the qubits sent from Alice to Bob as it does not contain anyinformation about the key as of yet

Eve knows about the usage of the third of the GHZ state but againshe needs qubits sent back to Alice from Bob for any legit informationabout the key.



When Bob sends the qubits back to Alice, Eve can catch the qubits.But measuring them is useless, as the density matrix of the qubits is

ρ = Tr |∆ >

= Tr12{

1√2

(|000 > +|111 >

)1√2

(< 000|+ < 111|

)}=

1

2

(|0 > +|1 >

) (< 0|+ < 1|

)or

ρ = Tr |∆′>

= Tr12{

1√2

(|001 > +|110 >

)1√2

(< 001|+ < 110|

)}=

1

2

(|0 > +|1 >

) (< 0|+ < 1|

)where Tr12 denotes the partial trace over qubit 1 and qubit 2.



Let’s consider the scenario case by case.

Case 1: Bob does nothing.Consider the qubit on which Bob does nothing, the state of thetripartite system is |∆ >. If Eve's measurement outcome is ’0’, thestate of the tripartite system will be |000 >. According to theprotocol Alice performs CNOT operation,

CNOT : |000 >→ |000 >

It can represented as

|000 >= |0 > ⊗ 1√2

(|Φ+ > +Φ− >

)Now its easy to see when Alice does the measurement on the last twoqubits, the outcomes are |Φ+ > and |Φ− > with equal probability 0.5



If Eve’s outcome is ’1’ this implies that the state of the tripartitesystem will be |111 >. Now if Alice performs CNOT,

CNOT : |111 >→ |011 >

And by the similar reasoning stated in the previous case Alice will get|Ψ+ > and |Ψ− > with equal probability 0.5 by doing a Bell statemeasurement. But in the protocol we know that if no eavesdroppersexist, its impossible for Alice to obtain measurement outcomes |Ψ− >.Once Alice gets |Ψ− > she knows that someone has eavesdropped.



Similarly we can show this for other two cases also.

Another strategy of attack from Eve could involve Eve capturing thequbit Bob sends to Alice and then using it as a control bit for CNOToperation on an auxiliary bit |0 >E owned by Eve.

It can be easily shown that even in this case Alice can find aboutEve’s existence with a probability of 0.5

Therefore, in all the cases where Eve has somehow to intercept andgain information, Alice comes to know about Eve’s existence with aprobability of 0.5

If they share n = 1000 qubits, then the probability that Alice wouldnot know about Eve’s existence is ≈ (10)−300


No Signaling and Quantum Key Distribution

What happens if someone comes up with a way that allows us tomeasure quantum states without necessarily causing the disturbancesin quantum states on which existing security proofs rely i.e. APostquantum Observer?

Most of the current QKD protocols rely on this fact to securelytransfer a secret key.

In the next part we find how we can develop secure QKD protocolagainst general attacks by a postquantum eavesdropper limited onlyby the impossibility of superluminal signaling (having a speed greaterthan that of light).

We can no longer assume that quantum theory correctly predicts thetrade-off between the information that Eve can extract and thedisturbance she must necessarily cause.


A quantum protocol for secret bit distribution

In this protocol Alice and Bob share n = MN2 pairs of systems, eachin the maximally entangled state |Ψ− >= 1√

2(|01 > −|10 >)

Both Alice and Bob measure their qubit in a random basis from theset

S = {Xr}

for r = 0 to N-1

Xr = {cos rπ2N|0 > +sin

rπ

2N|1 >,−sin rπ

2N|0 > +cos

rπ

2N|1 >}

After the measurements are done, Alice and Bob announce theirbases in public and abort the protocol and restart unless

2MN ≤∑i=1

∑c=−1,0,1

|{j : Aj = Xi ,Bj = Xi+c}|



Abortion Condition:

2MN ≥∑i=1

∑c=−1,0,1

|{j : Aj = Xi ,Bj = Xi+c}|

The probability of the above condition failing is of the order e−MN6

Proof* :

RHS =2MN∑i=0

(MN2

i

)(3

N

)i (1− 3

N

)MN2−i

We approximate this binomial distribution by applying theMultiplicative form of Chernoff Bound which posits:

Pr(X ≤ (1− δ)µ) ≤ e−δ2µ2 , 0 ≤ δ ≤ 1

In our case µ = np = 3MN and δ = 13 , so we get

e−δ2µ2 = e−

MN6



The outcomes are kept secret for one randomly chosen pair for whichthe bases chosen were Xi and Xi+c for some i and c = −1, 0, or 1.The outcomes are announced for all the remaining pairs

Alice and Bob abort the protocol if their outcomes a and b are notanticorrelated (i .e a 6= b) in all the cases they chose neighboring oridentical bases.

If the protocol is not aborted, their unannounced outcomes define thesecret bit, which is taken by Alice to be equal to her outcome and byBob to be opposite to his.


References

Jonathan Barrett (2005)

No Signaling and Quantum Key Distribution

PRL 95, 010503 (2005)

Xiaoyu Li (2002)

A quantum key distribution protocol without classical communications

Institute of Computing Technology, Chinese Academy of Sciences P.O.Box 2704,Beijing, 100080, P.R.China

Antonio Acn (2006)

Efficient quantum key distribution secure against no-signalling eavesdroppers

2006 New J. Phys. 8 126


Thank you!


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